Video Transcript
In this video, we’re going to learn
about kinematics and forces. We’ll learn about objects in motion
described through the kinematic equations, and we’ll see how forces help to create
and explain this motion.
To start out, imagine that you’re
participating in a stunt at an airshow. You’re in a plane flying straight
ahead with a speed 𝑣 in line with a bull’s-eye target ahead on the ground. You want to find out what is the
horizontal distance along the ground from the center of the bullseye called 𝑑 at
which you can release a ball from the plane which will fall and land in the center
of the bull’s-eye.
You know the plane is an altitude
𝐴 above ground and that as it falls the ball will encounter a wind which pushes on
the ball with the force equal to half the ball’s weight. Given all this information, to
solve for 𝑑, we’ll want to know something about kinematics as well as forces.
Kinematics is a field that
describes object motion under constant acceleration. And when we think of object motion,
we can consider that to be an object’s acceleration, velocity, and displacement. To describe an object’s motion, we
have four equations, all of which are in terms of acceleration, velocity,
displacement, and time.
Looking over these four equations,
we see that two of them involve solving for 𝑣 sub 𝑓, an object’s final velocity,
and two involve solving for 𝑑, an object’s displacement. But the particular algebraic form
of each equation doesn’t limit what variable we solve for. For example, considering the
equation at the bottom left, we can rearrange this equation to solve for any of the
four variables involved, depending on the information we had and the information we
wanted to solve for. That’s true for all four of the
kinematic equations.
Considering these equations, the
question may come up of where they come from. These equations do have an
origin. They’re derived from the
definitions of the terms involved in the equations, acceleration, velocity, and
displacement.
For example, the definition of
acceleration 𝑎 is that it’s equal to the change in velocity over the change in
time. We can rewrite this expression to
say instead of Δ𝑣, that that change is equal to a final velocity minus an initial
velocity. Then, if we multiply both sides of
the equation by Δ𝑡, the time difference, that expression cancels on the right-hand
side. And if we add the initial velocity
𝑣 sub zero to both sides, that term also cancels on the right-hand side. We’re left with an expression that
says the final velocity is equal to the initial velocity plus 𝑎 times the change in
time.
This equation is the same as the
one written in the top left of our four. We won’t derive the other three
kinematic equations. But this example- we won’t derive
the other three kinematic equations, but all of them do come from these basic
definitions of acceleration, velocity, and displacement.
When using the kinematic equations
practically, we want to be sure of two things. First, the kinematic equations
apply only when acceleration is constant over whatever motion or process we’re
considering. Over any interval where
acceleration is not constant, that is it changes, these equations are not valid. Second, because of the
directionality involved in positive-negative velocity, acceleration, or
displacement, we want to be careful to establish those directions at the outset of a
problem and carefully follow them as we go through a solution. That’s a little bit about
kinematics. Now, let’s talk about how
kinematics work together with forces.
Forces connect with the kinematic
equations because forces, we know, create motion. In particular, when a force on an
object is unbalanced, that object accelerates. It’s Newton’s second law of motion
that tells us how an object accelerates under an unbalanced force. This second law is a vector
equation, meaning that the direction of the net force is the same as the direction
of the object’s acceleration. So long as the net force acting on
an object is constant, so is that object’s resulting acceleration, which is then the
acceleration that we use in our kinematic equations to solve for object motion.
As they’re written, these equations
of motion express magnitude quantities, speeds, acceleration magnitudes, and
distances. But we know that, in general,
acceleration, velocity, and displacement are vectors. In these equations, the vector
nature of the terms involved is preserved and maintained using positive and negative
signs. That’s part of why establishing a
sign convention in a kinematics and forces example is so important.
Putting this altogether, let’s
consider an approach for solving problems involving both forces and kinematics. First, if we’re not given the net
force acting in our particular scenario, a free body diagram can help us find
that. Once we have that net force, the
second law of motion can let us use it to solve for the object’s resulting
acceleration 𝑎.
Once we’ve collected all of the
motion-based information we can gather, we choose positive and negative directions
for those terms, velocity, acceleration, and so forth. Then, looking over the list of
kinematic equations, we choose the one or ones that will help us solve for what we
want to find. Let’s practice using this approach
through an example.
An object is acted on by three
simultaneous forces. 𝐅 one equals negative 3.00𝐢 plus
2.00𝐣 newtons, 𝐅 two equals 6.00𝐢 minus 4.00𝐣 newtons, and 𝐅 three equals
2.00𝐢 plus 5.00𝐣 newtons. The object experiences acceleration
of 4.23 meters per second squared. The object is initially at
rest. Find the mass of the object. Find the speed of the object after
5.00 seconds. Find the components of the velocity
of the object after 5.00 seconds.
Given the three forces 𝐅 one, 𝐅
two, and 𝐅 three acting on an object initially at rest, which result in an
acceleration of 4.23 meters per second squared, we want to solve first for the mass
of the object. We can label that 𝑚. We also want to solve for the speed
the object attains after 5.00 seconds. We’ll call this speed 𝑠. And finally, we want to solve for
the velocity components of the object after that same amount of time. We’ll call these components 𝑣 sub
𝑥 and 𝑣 sub 𝑦.
To start on our solution, we can
draw a sketch of the forces acting on this object. If we sketch in the three forces
acting on our object on a pair of axes in units of newtons, we see that we can add
these together algebraically to solve for the net force acting on the object. Writing out 𝐹 one, 𝐹 two, and 𝐹
three by their component parts, when we add them together, we find a net resultant
force of 5.00𝑖 plus 3.00𝑗 newtons.
Now that we know the net force, we
can use Newton’s second law, which says this net force equals the object’s mass
times its acceleration to solve for object mass 𝑚. By the second law, object mass is
equal to the magnitude of the net force acting on it divided by its acceleration
𝑎. We solve for the magnitude of the
net force by squaring each of its two components, adding the results together, and
taking the square root of that sum.
Dividing that magnitude by our
acceleration, given as 4.23 meters per second squared, we find that object mass, to
three significant figures, is 1.38 kilograms. That’s the mass of the object under
the influence of these three forces.
Next, we want to solve for the
speed of our object after a time of 5.00 seconds has elapsed. Since the acceleration, in this
case, is a constant value across that time, this tells us that the kinematic
equations of motion apply to this object’s motion. We look through these different
equations seeking a match between what we’re trying to solve for, final speed, and
the information we already know.
Since we’re told that time, which
we can call 𝑡, is 5.00 seconds, the first equation written down is a match for what
we want to solve for. Written in terms of our variables,
this equation is 𝑠 is equal to the initial speed of the object plus its
acceleration times time. We know that because it started at
rest, its initial speed is zero. So, its final speed is simply 𝑎
times 𝑡, or 4.23 meters per second squared times 5.00 seconds. This is equal to 21.2 meters per
second. That’s the object’s final speed
after 5.00 seconds under this acceleration.
Lastly, we want to solve for the
components of the velocity of the object at the same time 𝑡 equals 5.00
seconds. To start on that solution, let’s
draw the net force acting on our object in on our graph. Our net force, the sum of 𝐹 one,
𝐹 two, and 𝐹 three, is pointed at some angle, we can call it 𝜃, from the positive
𝑥-axis. This angle is important because the
direction of our net force by Newton’s second law is the same as the direction of
the resulting acceleration. And since our particle started from
rest, that means its velocity will also be in the direction of force since it’s in
the direction of acceleration.
Here’s what we know so far about
our final velocity at 𝑡 equals 5.00 seconds. We know, first of all, that its
magnitude is equal to 𝑠, which we’ve solved for. And second, we know that this
velocity has an 𝑥- and a 𝑦-component. But we just don’t know what those
components are. However, if we solve for the
direction of the net force, then we can figure out how the final speed magnitude
divides up over the two components of velocity.
So, looking at our diagram, if we
draw in the 𝑥- and 𝑦-components of the net force and recognize we know their
numerical values from solving for them earlier, we can write that the tangent of our
angle 𝜃 is equal to 3.00 newtons divided by 5.00 newtons. Taking the arctangent of both
sides, we now have an expression for the angle 𝜃. And when we solve for it, we find
it’s roughly equal to 31 degrees.
Now, that we know the angle 𝜃, we
can solve for the components of velocity based on the final magnitude of that
velocity, which is speed. We can write 𝑣 sub 𝑥 as the
magnitude of the velocity, which is 𝑠, times the cos of 𝜃 and 𝑣 sub 𝑦 as 𝑠
times the sin of 𝜃. Plugging in for these values when
we solve for 𝑣 sub 𝑥 and 𝑣 sub 𝑦, we find they’re equal, to three significant
figures, to 18.1 meters per second and 10.9 meters per second, respectively. In component form then, our final
velocity is 18.1𝑖 plus 10.9𝑗 meters per second.
Let’s take a minute to review what
we’ve learnt about kinematics and forces. We’ve seen here that force
imbalances are what accelerate objects and that acceleration relationship is found
through Newton’s second law. We’ve also seen that when an
object’s acceleration is constant, its motion can then be described by what are
called the kinematic equations. These four equations are based on
the relationships between acceleration, velocity, and displacement.
We’ve also gotten practice using a
problem-solving strategy for examples with kinematics and forces. First, we’ve solved for the net
force involved in a scenario. Then, we apply Newton’s second law
of motion to get information about its acceleration. Then, we established which
directions are in the positive and which directions are in the negative. And finally, we apply the kinematic
equation or equations that let us solve for the variable we’re seeking. Knowing this process and these four
equations are helpful for understanding kinematics and forces.
